Math help - eigenvalues

[RESmonkey]

Find all eigenvalues (lambda symbol) (set of symbol) R and eigenfunctions y(x) for the eigenvalue problem

y" + [lamba symbol]*y = 0, 0 < x < Pi/2

with the mixed boundary conditions y'(0) = 0 , y(Pi/2) = 0 .

Hint First consider [lambda symbol] = 0, then [lambda symbol] < 0, then [lambda symbol] > 0.






^^^^ Ok, I need help. Second order ODE I'm guessing from the looks of it. I only know how to find eigenvalue and eigenvectors in things like ----> y' = [A]*y where [A] is a matrix.


Please help , I know there are a bunch of users who know this stuff

 

[Zim Hosein]

I :heart: :beer:

 

[Turin39789]

Originally posted by: Zim Hosein
I :heart: :beer:

sigh. I was going to be done, one more for bedtime

 

[MotF Bane]

Originally posted by: Zim Hosein
I :heart: :beer:

:thumbsup:

 

[LtPage1]

42
The Battle of Waterloo
December 7, 1941
Don't make me break out my googlestick and beat you with it.

 

[Fenixgoon]

all i remember from solving the wave equation (1D PDE) was a series of sines and cosines. sorry i'm useless

but if we're going off of simple functions (not matrices)

i wouldn't be surprised if the answers took the form of sine(x), cosine(x), exp(x), exp(-x)

 

[oogabooga]

Originally posted by: Zim Hosein
I :heart: :beer:

:heart:

 

[dotcom173]

i had linear analysis/diff equations last quarter, so i should know this, but i dont.

I want to say that you turn it into the 'characteristic equation' which you get by replacing y'' with r^2, and if there were a y', it would be replaced with r.

So, you get r^2 + Q = 0. Let's pretend Q is lambda.

Then solve for r: r^2 = -Q ----> r = Rad(-Q), Now that i get to this step, i realize i don't think im doing it right. I want to say you're answer for lambdas will consist of imaginary numbers with 'i'

EDIT: oo oo oo, i think then you put those lambda's into the 'general equation,' which is like:

C1e^Q1t + C2e^Q2t = 0, where C1 and C2 are constants you find when you plug in the initial values it gave you, and Q1 and Q2 are your eigen values, assuming there were exactly 2.

Were you not given a matrix? I just looked back in my book, and it seems like i was always given a matrix, then you put in Lambda into the matrix, and when you do row operations, you eventually get what the lambda's are

 

[Hacp]

f=c/lambda

 

[Turin39789]

Originally posted by: dotcom173
i had linear analysis/diff equations last quarter, so i should know this, but i dont.

I want to say that you turn it into the 'characteristic equation' which you get by replacing y'' with r^2, and if there were a y', it would be replaced with r.

So, you get r^2 + Q = 0. Let's pretend Q is lambda.

Then solve for r: r^2 = -Q ----> r = Rad(-Q), Now that i get to this step, i realize i don't think im doing it right. I want to say you're answer for lambdas will consist of imaginary numbers with 'i'

EDIT: oo oo oo, i think then you put those lambda's into the 'general equation,' which is like:

C1e^Q1t + C2e^Q2t = 0, where C1 and C2 are constants you find when you plug in the initial values it gave you, and Q1 and Q2 are your eigen values, assuming there were exactly 2.

Were you not given a matrix? I just looked back in my book, and it seems like i was always given a matrix, then you put in Lambda into the matrix, and when you do row operations, you eventually get what the lambda's are

diagonals minus lamba solve for =0 iirc

 

[dotcom173]

Originally posted by: Turin39789
diagonals minus lamba solve for =0 iirc

Yup, that's how i did it. Finding the evalues is usually the simple part, its finding the vectors associated with each evalue, and then putting all that shit into a fucking equation, god that class sucked...luckily im in linear 2 now!!! Yay, not.

 

[RESmonkey]

Thanks guys. Dotcom, you're thinking of the first order ones where they give you the matrix. These are boundary value problems, and I found some help/similar stuff here:


http://tutorial.math.lamar.edu/Classes/DE/BVPEvals.aspx
http://tutorial.math.lamar.edu/Classes/DE/DE.aspx

 

[silverpig]

I could probably figure it out, but I'm kind of lazy.

L = 0 should be fairly easy though.
L < 0 means that you have an i in there as part of whatever comes down from your differentiation
L > 0 means it's real.

Look for solutions of the type A*e^(f(x)) or something.

 

[interchange]

I have forgotten way too much math in my lifetime.

 

[Legendary]

Jebus my Math degree was apparently useless.
CompSci FTW.

 

[Unitary]

This is a pretty straight foward problem from PDE's: (Heat equation with one insulated end and another held at zero?)

Using the hint, if L=0 then the ODE becomes y''=0. The general solution is y(x)=a+bx. Now y'(x)=b, so the condition y'(x)=0 gives b=0. Similarly, if y(P/2)=0, then we get a=0. We can conclude that L=0 is not an eigenvalue since there are no nonzero solutions.

Now suppose that L<0. For convenience suppose that L=-c^2. Then the ODE reads y''-c^2y=0, which has general solution y(x)=ae^(-cx)+be^(cx). If you use the boundary conditions, you find that both the arbitrary constants a,b are zero. So, there are no negative eigenvalues since there are no nonzero solutions.

Lastly, check if L=c^2>0. Then the ODE is y''+c^2 y=0. The general solution is a Cos(cx)+b Sin(cx). The condition that y'(0)=0 gives you, b=0. Using that y(P/2)=0 we must have Cos(c* Pi/2)=0 This holds whenever c is a positive integer. c=1,2,3,4.... Since L=c^2, we get eigenvalues L=1,4,9,16,... with eigenfunctions y(x)=Cos(n x)

Hope this helps! The best thing to remember is the general method for solving these boundary value (eigenvalue) problems. Suppose L is either <0, =0 or >0 to help you find a general solution to work with, the determine if any non-zero solutions exist.

Best of luck!

 

[Cattlegod]

ahh diffeq, the wonderful math that you will never use ever in your life.

 

[TallBill]

We need a super genius

 

[RichieZ]

sorry this was like 6-7 years ago for me, but i'm sure i forgot all this stuff as soon that qtr finished.

good luckl!

 

[Unitary]

Originally posted by: Cattlegod
ahh diffeq, the wonderful math that you will never use ever in your life.

Too bad that you believe this... couldn't be farther from the truth.

 

[CraKaJaX]

Originally posted by: Cattlegod
ahh diffeq, the wonderful math that you will never use ever in your life.

Yep. I remember our prof gave us the equation for an eel and the equation for a snowflake in class one day. :roll: Taking linear algebra right now... it's actually really helpful. Diff EQ however, was not.

 

[Jeff7]

Oh god, no...terrible memories of Partial and Ordinary Differential Equations, Math 251, are returning.

....
Somebody help me! Mommy! I'm sorry I spilled the transmission fluid, mommy! No, no! Don't weld me to the wall, mommy!!!
Ehahaha!! AAAAHhhhhh!!!!




(Calculus of any sort was most unkind to me. 2+2, addition, algebra, geometry, trig - those all mean something to me. Calculus, derivatives, integrals - they were more like just memorizing, like might be done in an Organic Chemistry class.)

 

[Rockinacoustic]

Originally posted by: Jeff7

(Calculus of any sort was most unkind to me. 2+2, addition, algebra, geometry, trig - those all mean something to me. Calculus, derivatives, integrals - they were more like just memorizing, like might be done in an Organic Chemistry class.)

Except with Orgo there's usually more than one way to get you're desired product

 

[blinky8225]

Originally posted by: RESmonkey
Find all eigenvalues (lambda symbol) (set of symbol) R and eigenfunctions y(x) for the eigenvalue problem

y" + [lamba symbol]*y = 0, 0 < x < Pi/2

with the mixed boundary conditions y'(0) = 0 , y(Pi/2) = 0 .

Hint First consider [lambda symbol] = 0, then [lambda symbol] < 0, then [lambda symbol] > 0.






^^^^ Ok, I need help. Second order ODE I'm guessing from the looks of it. I only know how to find eigenvalue and eigenvectors in things like ----> y' = [A]*y where [A] is a matrix.


Please help , I know there are a bunch of users who know this stuff

Pretty simple actually. Just solve it like a standard second order differential equation.

Let [lamba symbol] = c^2

y'' + c^2 *y = 0

I'm going to assume you already know how to solve this, but if you can't I can show you how.

So you solve that and get:
y(x) = Acos(cx) + Bsin(cx) where A and B are constants.

Next let us deal with the boundary conditions. y'(x) = 0 and y(Pi/2) = 0.

y'(x) = 0 = -Ac * sin(0) + Bc*cos(0) = B*c ==> B = 0 since you don't want the trivial solution c = 0.

Since B = 0, y(x) = Acos(cx)

y(Pi/2) = 0 = Acos(c*Pi/2). We can have A = 0 or c*Pi/2 = 0.

We will have c*Pi/2 because A = 0 leaves us with only the trivial solution y = 0.

So for c*Pi/2 to be true, c = 1,2,3,4,5,6.........
your Eigenvalues are [lamba symbol]=1,4,9,16,...

Yes there are an infinite amount. As for the hint my substitution of [lamba symbol] = c^2 is simply saying [lambda symbol] > 0. If you try the other two cases, you will only find the trivial solution, y = 0.

Then, your eigenfunctions are y(x) = cos(n^2 * x), where n is any integer > 0.

 

[TecHNooB]

Originally posted by: Jeff7
Oh god, no...terrible memories of Partial and Ordinary Differential Equations, Math 251, are returning.

....
Somebody help me! Mommy! I'm sorry I spilled the transmission fluid, mommy! No, no! Don't weld me to the wall, mommy!!!
Ehahaha!! AAAAHhhhhh!!!!




(Calculus of any sort was most unkind to me. 2+2, addition, algebra, geometry, trig - those all mean something to me. Calculus, derivatives, integrals - they were more like just memorizing, like might be done in an Organic Chemistry class.)

That's so wrong. Calculus is amazing! Maybe some of the procedures may seem like memorizing if you have forgetten the proofs, but the results very useful!